|x+19|<5

Solution

∣ x + 19 ∣ < 5

- 24 < x < - 14

Interval Notation

(- 24, - 14)

Solution steps

∣ x + 19 ∣ < 5

Apply absolute rule: If

∣ u ∣ < a, a > 0

then

- a < u < a

- 5 < x + 19 < 5

x + 19 > - 5 and x + 19 < 5

x > - 24 and x < - 14

Merge Overlapping Intervals

- 24 < x < - 14

b^{4} (3 a b^{3}) (2 a^{2} b^{- 5})

119 - 0.16 \cdot 28 \cdot 9.81 = 28 \cdot x

x^{2} + (6 x + 22)^{2} = (6 x + 23)^{2}

100 x^{4} - 60 x^{2} z + 9 z^{2}

\frac{2 x ^{3} y ^{4}}{x ^{5} y ^{3}}